Symmetric functions and roots of polynomials

Question

$f = x^3-\frac{1}{2}x^2+1$ and their roots $a,b,c$. I want to find polynomial of degree 3 with roots $a^4,b^4,c^4$. I know that i need express $e_i(a^4,b^4,c^4)$ in terms of $e_i(a,b,c)$ those coefficients of f. But if i do it directly it will be time-consumming process. Is there much easier way?

Answer 1

Doing $(a,b,c)→(a^2,b^2,c^2)$ two times is clearly the easier way.
Let $A=a+b+c,B=ab+bc+ca,C=abc$.
Consider $$\begin{align}&(x^{1/2}-a)(x^{1/2}-b)(x^{1/2}-c)\times (x^{1/2}+a)(x^{1/2}+b)(x^{1/2}+c) = \\&(x^{3/2}-Ax+Bx^{1/2}-C)\times (x^{3/2}+Ax+Bx^{1/2}+C) = \\&(x^{3/2}+Bx^{1/2})^2 - (Ax+C)^2 = \\&x^3+2Bx^2+B^2x-A^2x^2-2ACx-C^2\end{align}$$
So, $(A,B,C)\rightarrow(2B-A^2,B^2-2AC,-C^2)$ does $(a,b,c)\rightarrow (a^2,b^2,c^2)$

Answer 2

One way is the following: $$\begin{align}(x-a^4)(x-b^4)(x-c^4)&=x^3-\left(\sum\limits_{cyc} t^4\right)x^2+\left(\sum\limits_{cyc} \left(\frac{1}{t}\right)^4\right)x-1\tag 1\\ &=x^3-\frac12\left(\left(\sum\limits_{cyc} t^3\right)-1\right)x^2 +\frac12\left(\sum\limits_{cyc} \left(\frac{1}{t}\right)^2\right)x-1\tag2\\ &=x^3-\frac12\left(\frac12\left(\sum\limits_{cyc} t^2\right)-4\right)x^2 +\frac{1}2x-1\tag4\\ &=x^3-\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}\right)^2-4\right)x^2+\frac12x-1\tag 5\end{align} $$

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• Move from $(0)$ to $(1)$ $abc=-1$ and $\sum_{cyc}(xy)^4=\sum_{cyc}\frac{(xyz)^4}{z^4}=\sum_{cyc}\frac{1}{t^4}$
• Move from $(1)$ to $(2)$ used $t^4=\frac{1}{2}t^3-t$ and $\frac{1}{t^4}=\frac{1}{2t^2}-\frac{1}{t}$ avec $\sum_{cyc}t=\frac12$ and $\sum_{cyc}\frac{1}{t}=\frac{ab+bc+ac}{abc}=0$

• Move from $(2)$ to $(4)$ use $t^3=\frac{1}{2}t^2-1$ and note that $ab+bc+ac=0$ so the sum $\sum_{cyc}\frac{1}{t^2}$ is : $$(ab)^2+(bc)^2+(ca)^2=\underbrace{(ab+bc+ac)^2}_{=0}-2abc\underbrace{(a+b+c)}_{=\frac{1}{2}}=1$$

• Move from $(4)$ to $(5)$ note that $(a+b+c)^2=a^2+b^2+c^2$ ($ab+bc+ac=0$)

Answer 3

Using the power of the Dandelin-Gräffe iteration. For any polynomial $p$ with roots $z_i$, the polynomial $p^{(1)}(x)$ defined via $$ p^{(1)}(x^2)=p(x)·p(-x) $$ has the same degree as $p$ and roots $z_i^2$. Repeating this process, the polynomial $p^{(2)}(x)$ defined via $$ p^{(2)}(x^2)=p^{(1)}(x)·p^{(1)}(-x) $$ has still the same degree as $p$ and roots $z_i^4$.

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For the given polynomial $f(x)=x(x^2)−\frac12x^2+1$, this iteration gives \begin{alignat}{4} p^{(1)}(x)&=(−\tfrac12x+1)^2 - x(x)^2&&=-x^3+\tfrac14x^2+x+1 &&=-x(x^2+1)+1+\tfrac14 x^2\\ p^{(2)}(x)&=(1+\tfrac14 x)^2 - x(x+1)^2&&=-x^3-\tfrac{31}{16}x^2-\tfrac12x+1 \end{alignat}